Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

class Main {
    public static void main(String[] args) {
        Scanner stdIn = new Scanner(System.in);
        int N = stdIn.nextInt();
        double[] p = new double[N];
        for (int i=0; i<N; i++) p[i] = stdIn.nextDouble();
        double[][] DP= new double[N+1][N+1];
        for (int i=1; i<N+1; i++) {
            for (int j = 0; j < N + 1; j++)
                DP[i][j] = 0;
        }
        DP[0][0] = 1;
        for (int i=1; i<N+1; i++) {
            for (int j=0; j<i+1; j++) {
                if (j == 0) DP[i][0] = DP[i-1][0] * (1 - p[i-1]);
                else if (i == 1 && j == 1) DP[i][j] = p[0];
                else DP[i][j] = DP[i-1][j] * (1 - p[i-1]) + DP[i-1][j-1] * p[i-1];
            }
        }
        double sum = 0;
        for (int i=N/2+1; i<N+1; i++) sum += DP[N][i];
        System.out.println(sum);
    }
}